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I am trying to understand why the special case of this theorem holds:

Liouville Theorem Let $u$ be a subharmonic function on $\mathbb{C}$ such that \begin{eqnarray} \limsup_{z \to \infty} \frac{u(z)}{\log |z|} \leq 0. \end{eqnarray} Then $u$ is constant on $\mathbb{C}$. In particular, every subharmonic function on $\mathbb{C}$ which is bounded above must be constant.

So why does the following implication hold for subharmonic functions?: \begin{eqnarray} u \, \text{bounded above} \quad \Rightarrow \quad \limsup_{z \to \infty} \frac{u(z)}{\log |z|} \leq 0 \end{eqnarray}

Any help would be greatly appreciated!

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That implication holds for any function defined on $\Bbb C$. If $u(z) \le M$ for all $z \in \Bbb C$ then $$ \frac{u(z)}{\log |z|} \le \frac{M}{\log |z|} $$ for $|z| > 1$ and therefore (since $\log |z| \to \infty $) $$ \limsup_{z \to \infty} \frac{u(z)}{\log |z|} \le \limsup_{z \to \infty} \frac{M}{\log |z|} = 0 \, . $$

In the case of a harmonic function which is bounded above it follows that the given theorem of Liouville can be applied to conclude that $u$ is constant.

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